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3 years ago

How find \empirical formula

1 answer:

6 0

The empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound. A simple example of this concept is that the empirical formula of sulfur monoxide, or SO, would simply be SO, as is the empirical formula of disulfur dioxide, S₂O₂. Thus, sulfur monoxide and disulfur dioxide, both compounds of sulphur and oxygen, have the same empirical formula. However, their molecular formulas, which express the number of atoms in each molecule of a chemical compound, are not the same.

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A 150 g ball and a 240 g ball are connected by a 33-cm-long, massless, rigid rod. The balls rotate about their center of mass at

stira [4]

Answer:

v= 3.18 m/s

Explanation:

Given that

m= 150 g = 0.15 kg

M= 240 g = 0.24 kg

Angular speed ,ω = 150 rpm

The speed in rad/s

$\omega =\dfrac{2\pi N}{60}$

$\omega =\dfrac{2\pi \times 150}{60}$

ω = 15.7 rad/s

The distance of center of mass from 150 g

$r=\dfrac{150\times 0+240\times 33}{150+240}\ cm$

r= 20.30 cm

The speed of the mass 150 g

v= ω r

v= 20.30 x 15.7 cm/s

v= 318.71 cm/s

v= 3.18 m/s

6 0

3 years ago

Hunters velocity was 4.5 m/s. At the end of his race, his velocity was the same. Which best describes his Movement?

hjlf

He has no acceleration because his velocity remains constant

8 0

3 years ago

When a solid compound dissolves in water,

rewona [7]

The answer is d. Each of its particles becomes surrounded by water molecules.

Notice that when you dissolve a solid compound in the water, it is no longer visible in or eyes, it mixes in the water, it is because it is being divided into smaller particles and each particles were surrounded by the water molecules.

6 0

3 years ago

A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is

ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0$

Solving the above equation we get

$t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89$